●Let us now see how randomization is done when a collision occurs.After a(71),time is divided into discrete slots whose length is equal to the worst-case round-trip propagation time on the ether(2τ). To accommodate the longest path allowed by Ethernet, the slot time has been set t0 512 bit times, or 51.2μsec.
After the first collision, each station waits either 0 or l (72) times before trying again. If two stations collide and each one picks the same random number, they will collide again. After the second collision, each one picks either 0,1,2,or 3 at random and waits that number of slot times. If a third collision occurs (the probability of this happening is 0.25), then the next time the number of slots to wait is chosen at (73) from the interval 0 to 23-1.
In general, after i collisions,a random number between 0 and 2i-1 is chosen, and that number of slots is skipped. However, after ten collisions have been reached, the randomization (74) is frozen at a maximum of 1023 slots. After 16 collisions, the controller throws in the towel and reports failure back to the computer. Further recoveryis up to (75 )layers.
(71) A.datagram
B.collision
c.connection
D. service
(72) A.slot
B.switch
C.process
D.fire
(73) A.rest
B.random
C.once
D.odds
(74) A.unicast
B.multicast
C.broadcast
D.interval
(75) A.local
B.next
C.higher
D.lower
第1题:
5W2H法要求人们从以下7个方面进行设问:What、Who、、When、Where、How、Howmuch。
Why
略
第2题:
使用哪个函数可插入当前日期?
A.today()
B.now()
C.time()
D.new()
第3题:
5W2H分析法,包括7个要素:why, what, who, when, where, how, how much.
第4题:
计划的主要内容包括______计划必须清楚的确定和描述这些内容。
A.What,Why,Who,Where,When,How many,How much
B.What,Why,We,Where,When,How to,How much
C.What,Why,We,Where,When,How,How much
D.What,Why,Who,Where,When,How,How much
第5题:
计划的主要内容包括______,计划必须清楚地确定和描述这些内容。
A.What, Why, Who, Where, When, How many, How much
B.What, Why, We, Where, When, How to, How much
C.What, Why, We, Where, When, How, How much
D.What, Why, Who, Where, When, How, How much
第6题:
3、5W2H分析法,包括7个要素:why, what, who, when, where, how, how much.